implicit differnetiation word problem

[Dorian Gray]

Greetings Mathematicians,

I was wondering if somebody could please look at my work for this problem. As you can tell from my previous posts, I am not the most comfortable with these kinds of problems. Here is the problem:

The formula for the volume of a cone is V=(1/3)(pi)(r squared)(h).Find the rate of change of volume if dr/dt = 2 in/min and h=3r when r=6 inches.

I attached a copy of my work. My answer seems rather high, which is why I am suspicious of my answer. Here is a picture of my work.




Any and all help, comments, suggestions, and advice is/are always appreciated.

 

[galactus]

$\displaystyle V=\frac{\pi}{3} r^{2}h$

We are told that h=3r, so we have:

$\displaystyle V=\frac{\pi}{3} r^{2}(3r)=\pi r^{3}$

$\displaystyle \frac{dV}{dt}=3\pi r^{2}\frac{dr}{dt}$

$\displaystyle \frac{dV}{dt}=3\pi (6)^{2}\cdot 2=216\pi$

 

[HallsofIvy]

Your error is only taking the derivative with respect to r when both r and h are changing!

The simplest way to handle this is just what galactus did- since h= 3r (for all r), replace h with that to get $\displaystyle V= \frac{\pi}{3}r^2h= \frac{\pi}{3}r^2(3r)= \pi r^3$

Another way is to use the product rule: since $\displaystyle V= \frac{\pi}{3}r^2h$, $\displaystyle \frac{dV}{dt}= \frac{\pi}{3}\frac{d(r^2h)}{dt}= \frac{\pi}{3}\left(2rh \frac{dr}{dt}+ r^2\frac{dh}{dt}\right)$.

Of course, you are told that r= 6 inches and dr/dt= 2 in/min. Since you are also told that h= 3r, h= 3(6)= 18 inches and dh/dt= 3(2)= 6 in/min.

 

[Dorian Gray]

Thank you very much!

Thank you both! I can clearly see why both of you did the things that you did.